Consider the following intermediate chemical equations:
[tex]
\begin{array}{ll}
NO(g)+O_3(g) \rightarrow NO_2(g)+O_2(g) & \Delta H_1=-198.9 kJ \\
\frac{3}{2} O_2(g) \rightarrow O_3(g) & \Delta H_2=142.3 kJ \\
O(g) \rightarrow \frac{1}{2} O_2(g) & \Delta H_3=-247.5 kJ
\end{array}
[/tex]
What is the enthalpy of the overall chemical equation [tex]NO ( g )+ O ( g ) \rightarrow NO _2(g)[/tex] ?
A. -305 kJ
B. -304.1 kJ
C. -93.7 kJ
D. 5887 k . l
Answer (2)
Reverse equations 2 and 3.
Add the reversed equations to equation 1.
Calculate the overall enthalpy change by summing the individual enthalpy changes: Δ H = − 198.9 − 142.3 + 247.5 .
The enthalpy of the overall chemical equation is − 93.7 k J .
Explanation
Understanding the Problem We are given three chemical equations with their corresponding enthalpy changes, and we want to find the enthalpy change for the overall reaction NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) . We can use Hess's Law, which states that the enthalpy change for a reaction is independent of the pathway taken, to find the enthalpy change for the overall reaction by manipulating the given equations.
Listing Given Equations First, we write down the given equations and their enthalpy changes:
NO ( g ) + O 3 ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) \t Δ H 1 = − 198.9 k J
2 3 O 2 ( g ) i g h t ha r p oo n u p O 3 ( g ) \t Δ H 2 = 142.3 k J
O ( g ) i g h t ha r p oo n u p 2 1 O 2 ( g ) \t Δ H 3 = − 247.5 k J
We want to obtain the reaction NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) .
Reversing Equations We need to manipulate the given equations to match the target equation. We can reverse equation (2) and equation (3) and then add them to equation (1).
Reverse equation (2): O 3 ( g ) i g h t ha r p oo n u p 2 3 O 2 ( g ) \t Δ H = − 142.3 k J
Reverse equation (3): 2 1 O 2 ( g ) i g h t ha r p oo n u pO ( g ) \t Δ H = 247.5 k J
Adding the Equations Now, we add equation (1), the reversed equation (2), and the reversed equation (3):
NO ( g ) + O 3 ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) \t Δ H 1 = − 198.9 k J O 3 ( g ) i g h t ha r p oo n u p 2 3 O 2 ( g ) \t Δ H = − 142.3 k J 2 1 O 2 ( g ) i g h t ha r p oo n u pO ( g ) \t Δ H = 247.5 k J
Adding these equations gives: NO ( g ) + O 3 ( g ) + 2 1 O 2 ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) + O ( g ) + 2 3 O 2 ( g )
Simplifying, we get: NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g )
Calculating Enthalpy Change To find the enthalpy change for the overall reaction, we add the enthalpy changes of the manipulated equations:
Δ H = − 198.9 + ( − 142.3 ) + 247.5 = − 198.9 − 142.3 + 247.5 = − 93.7 k J
Final Answer Therefore, the enthalpy change for the overall reaction NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) is − 93.7 k J .
Examples
Hess's Law is useful in determining the enthalpy change of a reaction that cannot be measured directly. For example, if you want to find the enthalpy change for the formation of a compound from its elements, but the reaction is too slow or produces unwanted side products, you can use Hess's Law to calculate the enthalpy change from a series of reactions that can be measured. This is commonly used in industrial chemistry to optimize reaction conditions and improve efficiency.
The enthalpy change for the reaction NO ( g ) + O ( g ) → N O 2 ( g ) is − 93.7 kJ , calculated using Hess's Law by manipulating the given equations. The correct answer is therefore option C: -93.7 kJ.
;
